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Khách
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TM
0
PK
1
NH
2
1 tháng 7 2017
... \(=\left(sin^2a\right)^2+2\cdot sin^2a\cdot cos^2+\left(cos^2a\right)^2=\left(sin^2a+cos^2a\right)^2=1^2=1\)
17 tháng 8 2019
a.\(1-\sin^2\alpha=\cos^2\alpha\)
b.\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
c.\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\sin^2\alpha\)
d.\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)
e.\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha=\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha.\cos^2\alpha=\sin^2\alpha\)
g.\(\cos^2\alpha+\cos^2\alpha.\tan^2\alpha=\cos^2\alpha\left(1+\tan^2\alpha\right)=\cos^2\alpha.\frac{1}{\cos^2\alpha}=1\)
29 tháng 10 2018
a) 1- \(sin^2\alpha\)= \(cos^2\alpha\)
b) (\(1-cos\alpha\))(\(1+cos\alpha\)) = 1 - cos2\(\alpha\) = sin2\(\alpha\)
c) 1 + cos2\(\alpha\) + sin2\(\alpha\) = \(1+1=2\)
d) sin\(\alpha\) - sin\(\alpha.cos^2\alpha\)
= \(sin\alpha\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
e) \(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha\)
= \(\left(sin^2\alpha\right)^2+2sin^2\alpha.cos^2\alpha+\left(cos^2\alpha\right)^2\)
= \(\left(sin^2\alpha+cos^2\alpha\right)^2=1^2=1\)
f) \(tan^2\alpha-sin^2\alpha.tan^2\alpha\)
= \(tan^2\alpha\left(1-sin^2\alpha\right)=tan^2\alpha.cos^2\alpha=sin^2\alpha\)
g) \(cos^2\alpha+tan^2\alpha.cos^2\alpha\)
= \(cos^2\alpha\left(1+tan^2\alpha\right)=cos^2\alpha.\dfrac{1}{cos^2\alpha}=1\)
h) \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)
= \(tan^2\alpha\left[cos^2\alpha+\left(cos^2\alpha+sin^2\alpha\right)-1\right]\)
= \(tan^2\alpha\left(cos^2\alpha+1-1\right)\)
= \(tan^2\alpha.cos^2\alpha=sin^2\alpha\)
HM
1
30 tháng 7 2018
=\(\frac{1-cos2a}{1+cos2a}\)\(\left(1+cos2a+\frac{1-cos2a}{2}-1\right)\)+\(\frac{1+cos2a}{2}\)
=\(\frac{1-cos2a}{1+cos2a}\)\(\left(cos2a+\frac{1-cos2a}{2}\right)\)+\(\frac{1+cos2a}{2}\)
=\(\frac{1-cos2a}{1+cos2a}\)\(\left(\frac{2cos2a+1-cos2a}{2}\right)\)+\(\frac{1+cos2a}{2}\)
=\(\frac{1-cos2a}{1+cos2a}\)\(\left(\frac{1+cos2a}{2}\right)\)+\(\frac{1+cos2a}{2}\)
=\(\frac{1-cos2a}{2}\)+\(\frac{1+cos2a}{2}\)
=\(\frac{1-cos2a+1+cos2a}{2}\)
=\(\frac{2}{2}\)=1
LM
27 tháng 9 2018
a) 1 + tan22 a =1 +(\(\dfrac{sina}{cosa}\))2 =\(\dfrac{sina+cosa}{cos^2a}\)=\(\dfrac{1}{cos^2a}\)
b) 1 + cot2 a= 1 +(\(\dfrac{cosa}{sina}\))2 = \(\dfrac{cosa+sina}{sin^2a}\)=\(\dfrac{1}{sin^2a}\)
c) tan2 a (2 sin2a + 3 cos2 a - 2)
=tan2 a[cos2 a +2 (\(sina^2+cos^2a\))-2 ]
=\(\dfrac{sin^2a}{cos^2a}\)×\(cos^2a=sin^2a\)
7 tháng 9 2022
b: \(1+cot^2a=1+\left(\dfrac{cosa}{sina}\right)^2=\dfrac{1}{sin^2a}\)
c: \(=tan^2a\left[2\left(1-cos^2a\right)+3cos^2a-2\right]\)
\(=tan^2a\left[cos^2a\right]\)
\(=\dfrac{sin^2a}{cos^2a}\cdot cos^2a=sin^2a\)
V
2
QA
17 tháng 12 2019
\(B=\sin^230^0+\sin^240^0+\sin^250^0+\sin^260^0\)
\(B=\sin^230^0+\sin^240^0+\cos^2\left(90^0-50^0\right)+\cos^2\left(90^0-60^0\right)\)
\(B=\sin^230^0+\sin^240^0+\cos^240^0+\cos^230^0\)
\(B=\left(\sin^230^0+\cos^230^0\right)\left(\sin^240^0+\cos^240^0\right)\)
\(B=1+1\)
\(B=2\)
Chúc bạn hok tốt!!! vvvvvvvv
H
17 tháng 12 2019
Ta có :\(\sin\left(60\right)=\cos\left(30\right)\)(phụ nhau)
\(\Leftrightarrow sin^2\left(60\right)=\cos^2\left(30\right)\)
và :\(sin^2\left(50\right)=\cos^2\left(40\right)\)(tương tự như trên nha bạn)
Thay vào biểu thức B ta có :
\(B=\sin^2\left(30\right)+sin^2\left(40\right)+\cos^2\left(30\right)+\cos^2\left(40\right)\)
\(B=1+1\)
\(B=2\)
chúc bạn học tốt :)
\(A=sin\alpha-sin\alpha\cdot cos^2\alpha\)
\(A=sin\alpha\left(1-cos^2\alpha\right)\)
\(A=sin\alpha\cdot sin^2\alpha\)
\(A=sin^3\alpha\)
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